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Trigonometry – I — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – I1 Mark
MCQ
$\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}$ is equal to
  • ✓
    $2 \operatorname{cosec} A$
  • B
    $2 \sec A$
  • C
    $2 \sin \mathrm{A}$
  • D
    $2 \cos A$
✓

Answer

Correct option: A.
$2 \operatorname{cosec} A$
$2 \operatorname{cosec} A$
Explanation:
$\frac{\tan A}{1+\sec A}+\frac{1+\sec A}{\tan A}$
$= \frac{\tan ^2 A+1+\sec ^2 \mathrm{~A}+2 \sec \mathrm{A}}{(1+\sec \mathrm{A}) \tan \mathrm{A}} $
$=\frac{\sec ^2 \mathrm{~A}+\sec ^2 \mathrm{~A}+2 \sec \mathrm{A}}{(1+\sec \mathrm{A}) \tan \mathrm{A}} \ldots\left[\because 1+\tan ^2 \mathrm{~A}=\sec ^2 \mathrm{~A}\right] $
$= \frac{2 \sec \mathrm{A}(\sec \mathrm{A}+1)}{(1+\sec \mathrm{A}) \tan \mathrm{A}}=\frac{2 \sec \mathrm{A}}{\tan \mathrm{A}} $
$= \frac{2}{\sin \mathrm{A}}=2 \operatorname{cosec} \mathrm{A}$

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